Inequalities

But I don't see how I can solve it from there. I'd just like to know the method please, so I can do it myself thank you.

First, you need to find the values of that make the function zero. Since you've factorised it, that's now very easy: and .

Next, imagine the number line with these numbers and marked, and the number moving from left to right along the line. When it's right over on the left-hand side, . When it's somewhere in the middle , and when it's going off at the right-hand end, . So, look at the signs of your factors and in each of these three ranges of values of . For instance:

When (e.g. ), is negative and is also negative. And a negative times a negative equals a positive. So is positive. Which is not what we want: we want values of that make .

Now look at the other two ranges: and , and see what happens to the signs in each case.

Thanks for all the help everyone. I totally understand this now. As for Pankaj's questions:

(i) and

(ii)

I hope they're right, and thanks for the questions.

(i)There appears to be a typo .In place of -2 you should have written 2.

(ii)There is a very simple way out to do such questions.Plot all the numbers on the number line where the factors become zero.Here we see that -4,2,7,11 are the numbers.
The number line gets split into 5 parts/intervals.
(-) .........(+) .........(+)............... (-)................ (+)
--------------------------------------------------------------Number line
........-4 ..........2....... .......7 ...............11...........

Starting from the rightmost interval start putting the (+) and (-) signs alternatively.But remember sign will not change across that number whose corresponding factor has even power,2 in this case.And you get the desired intervals which are
Remember to exclude 2 since at x=2 LHS becomes 0.Your initial question can be answered in the same manner